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# Tutor profile: Noor S.

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Noor S.
Mathematics Teacher with 9 years Experience
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## Questions

### Subject:Trigonometry

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Question:

Solve for x ( 0 < x < 90 ) : 2 cos 3x - 1 = 0

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Noor S.
Answer:

Solution : 2 cos 3 x - 1 = 0 2 cos 3x = 1 ( took the 1 to the right hand side ) cos 3x = 1 /2 ( when 2 comes to the right hand side it will become in divided form) cos 3x = cos 60 ( according to trigonometric table cos 60 = 1/2 ) 3x = 60 ( cancel the cos in both sides ) x = 60 /3 ( when 3 comes to the right hand side it will become in divided form ) x = 20 ( according to condition ( 0 < x < 90 ) x value GREATER to ZERO SMALLER to 90 )

### Subject:Algebra

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Question:

solve the Quadratic Equation Factories : a^2 + 10a + 24

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Noor S.
Answer:

solution : a^2 + 10 a + 24 product = 24 a^2 + 6a + 4a + 24 = 0 Middle = + 10 a *a + 6 *a + 4 *a + 6 *4 = 0 Factor = 6 * 4 = 24 a ( a + 6 ) + 4 ( a + 6 ) = 0 Middle = + 6 + 4 = 10 ( a + 6 ) ( a + 4 ) = 0 ( a + 6 ) = 0 or ( a + 4 ) = 0 a = - 6 or a = - 4

### Subject:Differential Equations

TutorMe
Question:

solve the differential equation y = e^x + 1 , and show that y^'' - y^' = 0

Inactive
Noor S.
Answer:

y = e^x + 1 differentiating both sides of this equation with respect to x , we get d (y) / dx = d (e^x + 1) / dx y^' = e^x (1) ( according to differential formula ) Now differentiate equation ( 1 ) with respect to x , we get d (y^') / dx = d (e^x) / dx y^" = e^x ( according to differential formula ) substituting the values of y^' and y ^" in the given differential equation we get L.H. S as L.H.S y^" - y^' = e^x - e^x = 0 L.H.S = R.H S Thus , the given function is the solution of the corresponding differential equation

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